average value a retarding force, symbolized by the dashpot in the figure, slows the motion of the weighted…

Explanation:

Step1: Recall the formula for the average value of a function

The average value of a function (y = f(t)) over the interval ([a,b]) is given by (\frac{1}{b - a}\int_{a}^{b}f(t)dt). Here, (a = 0), (b=2\pi), and (f(t)=2e^{-t}\cos t). So, the average value (y_{avg}=\frac{1}{2\pi-0}\int_{0}^{2\pi}2e^{-t}\cos tdt=\frac{1}{\pi}\int_{0}^{2\pi}e^{-t}\cos tdt).

Step2: Use integration by parts

Let (u = \cos t), (dv=e^{-t}dt). Then (du=-\sin tdt), (v=-e^{-t}). By the integration - by - parts formula (\int_{}^{}u;dv=uv-\int_{}^{}v;du), we have (\int e^{-t}\cos tdt=-e^{-t}\cos t-\int e^{-t}\sin tdt). For (\int e^{-t}\sin tdt), use integration by parts again. Let (u = \sin t), (dv = e^{-t}dt). Then (du=\cos tdt), (v=-e^{-t}). So (\int e^{-t}\sin tdt=-e^{-t}\sin t+\int e^{-t}\cos tdt).

Step3: Solve the integral equation

Substitute the second integration - by - parts result into the first one: (\int e^{-t}\cos tdt=-e^{-t}\cos t-(-e^{-t}\sin t+\int e^{-t}\cos tdt)) (\int e^{-t}\cos tdt=-e^{-t}\cos t + e^{-t}\sin t-\int e^{-t}\cos tdt) (2\int e^{-t}\cos tdt=-e^{-t}\cos t + e^{-t}\sin t + C) (\int e^{-t}\cos tdt=\frac{1}{2}e^{-t}(\sin t-\cos t)+C)

Step4: Evaluate the definite integral

(\frac{1}{\pi}\int_{0}^{2\pi}e^{-t}\cos tdt=\frac{1}{2\pi}[e^{-t}(\sin t-\cos t)]_{0}^{2\pi}) When (t = 2\pi), (e^{-2\pi}(\sin(2\pi)-\cos(2\pi))=-e^{-2\pi}) When (t = 0), (e^{0}(\sin(0)-\cos(0))=- 1) (\frac{1}{2\pi}(-e^{-2\pi}+1))

Answer:

(\frac{1 - e^{-2\pi}}{2\pi})